sanchez (es25396) – Homework #3  Accelerated motion/ Free fall – guevara – (130101)
1
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16
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001
(part 1 of 6) 10.0 points
Consider the plot below describing motion of
an object along a straight path as shown in
the figure below.

2
0
2
4
6
0
2
4
6
8
10 12 14 16 18 20
time (s)
velocity (m/s)
Find the average acceleration during the
time interval 0 s to 5 s.
Correct answer: 0 m
/
s
2
.
Explanation:
Given :
v
i
= 5 m
/
s
,
v
f
= 5 m
/
s
,
t
i
= 0 s
and
t
f
= 5 s
.
a
avg
=
Δ
v
Δ
t
=
v
f

v
i
Δ
t
=
(5 m
/
s)

(5 m
/
s)
(5 s)

(0 s)
=
0 m
/
s
2
.
002
(part 2 of 6) 10.0 points
Find the average acceleration during the time
interval 5 s to 15 s.
Correct answer:

0
.
4 m
/
s
2
.
Explanation:
Given :
v
i
= 5 m
/
s
,
v
f
= 1 m
/
s
,
t
i
= 5 s
and
t
f
= 15 s
.
a
avg
=
(1 m
/
s)

(5 m
/
s)
(15 s)

(5 s)
=

0
.
4 m
/
s
2
.
003
(part 3 of 6) 10.0 points
Find the average acceleration during the time
interval 0 s to 20 s.
Correct answer:

0
.
2 m
/
s
2
.
Explanation:
Given :
v
i
= 5 m
/
s
,
v
f
= 1 m
/
s
,
t
i
= 0 s
and
t
f
= 20 s
.
a
avg
=
(1 m
/
s)

(5 m
/
s)
(20 s)

(0 s)
=

0
.
2 m
/
s
2
.
004
(part 4 of 6) 10.0 points
Find the instantaneous acceleration at 2 s.
Correct answer: 0 m
/
s
2
.
Explanation:
At
t
= 2 s, the slope of the tangent line is
zero; therefore the acceleration is
zero
.
005
(part 5 of 6) 10.0 points
Find the instantaneous acceleration at 10 s.
Correct answer:

0
.
4 m
/
s
2
.
Explanation:
At
t
= 10 s, the slope of the tangent line is
the same as in Part 2.
a
avg
=
(1 m
/
s)

(5 m
/
s)
(15 s)

(5 s)
=

0
.
4 m
/
s
2
.
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sanchez (es25396) – Homework #3  Accelerated motion/ Free fall – guevara – (130101)
2
006
(part 6 of 6) 10.0 points
Find the instantaneous acceleration at 18 s.
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 Spring '09
 Guevara
 Acceleration, Work, Correct Answer, m/s, Sanchez, Aavg

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